Question

In a dairy processing plant, milk enters a 30 m long and 2 cm diameter tube at 60 °C and leaves at 57 °C. The total heat loss over the tube length is 381.15 W. The specific heat capacity, density, and viscosity of milk are 3.85 kJ kg\(^{-1}\) K\(^{-1}\), 1020 kg m\(^{-3}\), and 1.20 cP, respectively. The Reynolds number for the flow is ________ (round off to the nearest integer).

Hint:

The Reynolds number is a dimensionless number used to predict flow patterns in different fluid flow situations. It is calculated using fluid velocity, pipe diameter, fluid density, and viscosity.

Answer 1

Correct Answer: 1700

We are given the following values:
- Length of the tube \( L = 30 \, \text{m} \),
- Diameter of the tube \( D = 2 \, \text{cm} = 0.02 \, \text{m} \),
- Temperature difference \( \Delta T = 60 - 57 = 3 \, \text{°C} \),
- Heat loss \( Q = 381.15 \, \text{W} \),
- Specific heat capacity \( c_p = 3.85 \, \text{kJ/kg K} = 3850 \, \text{J/kg K} \),
- Density \( \rho = 1020 \, \text{kg/m}^3 \),
- Viscosity \( \mu = 1.20 \, \text{cP} = 1.20 \times 10^{-3} \, \text{Pa.s} \).
The heat transfer equation for the tube is given by: \[ Q = \frac{\pi D L \rho c_p \Delta T}{4} \] Rearranging to solve for the mass flow rate \( \dot{m} \): \[ \dot{m} = \frac{4Q}{\pi D L \rho c_p \Delta T} \] Substituting the values: \[ \dot{m} = \frac{4 \times 381.15}{\pi \times 0.02 \times 30 \times 1020 \times 3850 \times 3} \] \[ \dot{m} = 0.022 \, \text{kg/s}. \] Now, we can calculate the Reynolds number (\( Re \)) using the formula: \[ Re = \frac{\rho v D}{\mu} \] To find \( v \) (the velocity of the flow), we use the mass flow rate and the density: \[ v = \frac{\dot{m}}{\rho A} = \frac{0.022}{1020 \times \pi \left(\frac{0.02}{2}\right)^2} \] \[ v = 0.173 \, \text{m/s}. \] Finally, substituting the values in the Reynolds number equation: \[ Re = \frac{1020 \times 0.173 \times 0.02}{1.20 \times 10^{-3}} = 1747.5. \] Thus, the Reynolds number is approximately \( 1750 \).